## Slot -2 – Quantitative Aptitude – Modern Maths – Progressions – Let a(base1), a(base2), …a(base52)

Let a(base1), a(base2), … , a(base52) be positive integers such that a(base1) < a(base2) < … < a(base52). Suppose, their arithmetic mean is one less than the arithmetic mean of a(base2), a(base3), …, a(base52). If a(base52) = 100, then the largest possible value of a(base1) is?

a) 45

b) 20

c) 48

d) 23

Solution:

We want to maximize the value of a1, subject to the condition that

(a2+ a3 +⋯….+ a52 )/51 -(a1+ a2 +⋯….+ a52 )/52 = 1

Since a(base52) = 100 and all the numbers are positive integers, maximizing a1 entails maximizing a2, a3 ….a51. The only way to do this is to assume that a2, a3…. A52 are in an AP with a common difference of 1.

Let the average of a2, a3, …, a52 i.e.

(a2+ a3 +⋯….+ a52 )/51= a27 =A ( using the average of an odd number of terms in an Arithmetic Progression is equal to the value of the middle-most term)

So a52  = a27 + 25*1 = a27 + 25  and

given a52 = 100

=>  a27 = A = 100 – 25 = 75

a2 + a3 + … + a52 = 75×51 = 3825

Given a1 + a2 +… + a52  = 52(A – 1) = 3848

Hence a1 = 3848 – 3825 = 23

Ans : 23

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## Slot -2 – Quantitative Aptitude – Modern Maths – Set Theory – If A = {6^2n -35n -1: n = 1,2,3,…}

If A = {6^2n -35n -1: n = 1,2,3,…} and B = {35(n-1) : n = 1,2,3,…} then which of the following is true?

a) Neither every member of A is in B nor every member of B is in A

b) Every member of A is in B and at least one member of B is not in A

c) Every member of B is in A.

d) At least one member of A is not in B

Answer: b) Every member of A is in B and at least one member of B is not in A.

Solution:
Given, A = 36^n – 35n – 1 = 36^n – 1^n – 35n

Since a^n – b^n is divisible by a – b for all positive integral values of n, So A is a

multiple of 35 for any integral value of n and B is a set containing all the

multiple of 35 including 0.

Hence, every member of A is in B but not every element of B is in A.

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## Slot -2 – Quantitative Aptitude – Modern Maths – Progressions – The arithmetic mean of x, y

The arithmetic mean of x, y and z is 80, and that of x, y, z, u and v is 75, where u=(x+y)/2 and v=(y+z)/2. If x ≥ z, then the minimum possible value of x is?

Solution:
Given, (x+y+z)/3=80

x+y+z=240——1)

And (x+y+z+u+v)/4=75

x+y+z+u+v =375——2)

From eq 1) & eq 2) u+v =135

And from question , u+v =(x+2y+z)/2

Or 270=x+2y+z

y = 30 and x+z = 210

as x ≥ z
so x will be minimum if x =z

minimum value of x =210/2 = 105

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## Slot -2 – Quantitative Aptitude – Modern Maths – Set Theory – For two sets A and B, let AΔB

For two sets A and B, let AΔB denote the set of elements which belong to A or B but not both. If P = {1,2,3,4}, Q = {2,3,5,6,}, R = {1,3,7,8,9}, S = {2,4,9,10}, then the number of elements in (PΔQ)Δ(RΔS) is?

a) 7

b) 8

c) 9

d) 6

Solution: Given, P = {1,2,3,4}, Q = {2,3,5,6,}, R = {1,3,7,8,9}, S = {2,4,9,10}

So (PΔQ) = { 1, 4, 5, 6} and (RΔS)= { 1,2,3,4, 7,8,10}

Thus (PΔQ)Δ(RΔS) = {2,3, 5,6, 7, 8,9,10}

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## Slot -2 – Quantitative Aptitude – Modern Maths – Sequence and Series – The value of the sum 7 x 11

The value of the sum 7 x 11 + 11 x 15 + 15 x 19 + …+ 95 x 99 is?

a) 80707

b) 80730

c) 80773

d) 80751

Solution: 7 x 11 + 11 x 15 + 15 x 19 + …+ 95 x 99

nth term of series = {4(n+1)-1}×{ 4 (n+1)+3}=16n^2+16+32n+8n+8-

3=16n^2+40n+21

Thus 7 x 11 + 11 x 15 + 15 x 19 + …+ 95 x 99 = ∑ Tn=∑ (16n^2+40n+15)

Required sum = ∑ Tn=∑ (16n^2+ 40n+15)=16 ×(n(n+1(2n+1))/6 + 40×n(n+1)/2 + 21n

Putting n= 23

Sum = 16×23×24×47/6+40×23×24/2+21×23

=69184+11040+483=80707

Correct option a) 80707

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## Slot -2 – Quantitative Aptitude – Modern Maths – P & C – How many two-digit numbers

How many two-digit numbers, with a non-zero digit in the units place, are there which are more than thrice the number formed by interchanging the positions of its digits?
a) 6

b) 8

c) 7

d) 5

Solution:

Let the number is ab so
10a+b >3(10b+a)

7a>29b

a/b>29/7

So the possible pair of (a,b) are (5,1), (6,1), (7,1) , (8,1) , (9,1) & (9,2)
Thus total 6 such numbers are possible.

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## Slot -2 – Quantitative Aptitude – Modern Maths – Sequence and Series – Let t1, t2,… be real numbers

Let t1, t2,… be real numbers such that t1+t2+…+tn = 2n2+9n+13, for every positive integer n ≥ 2. If tk=103, then k equals?

Solution:
Given, t(base1)+t(base2)+…+t(base n) = 2n^2+9n+13

So t(base1) + t(base2) = 2×2^2+9×2+13=39

t(base1) + t(base2) + t(base3) = 2×3^2+9×3+13=58 means t(base 3) = 58 – 39 = 19
t(base1) + t(base2) + t(base3)+t(base4) = 2×4^2+9×4+13=81 means t(base4) = 81 – 58 = 23
t(base1) + t(base2) + t(base3)+t(base4) + t(base5) = 2×5^2+9×5+13=108 means t(base4) = 108 – 81 = 27
t(base1) + t(base2) + t(base3)+t(base4) + t(base5) + t(base6) = 2×6^2+9×6+13=139 means t(base4) = 139 – 108 = 31

So we can see that t(base3), t(base4) , t(base5) , t(base6) form an A.P. 19, 23, 27, 31

t(basek) = 4(k+1) + 3

Given , t(basek) = 103
4(k+1) + 3 = 103
k = 24

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## Slot -2 – Quantitative Aptitude – Modern Maths – Permutation and Combination – In a tournament, there are 43 junior

In a tournament, there are 43 junior level and 51 senior level participants. Each pair of juniors play one match. Each pair of seniors play one match. There is no junior versus senior match. The number of girl versus girl matches in junior level is 153, while the number of boy versus boy matches in senior level is 276. The number of matches a boy plays against a girl is?

Solution: total number of matches = 43C2 + 51C2 = 2178
Girl vs girl matches at junior level = 153 = (17×18)/2

Thus number of girls = 18
So number of boys at junior level = 43 – 18 = 25

the number of boy versus boy matches in senior level = 276 =(23×24)/2

Number of boys at senior level = 24
Number of girls at senior level = 51 – 24 = 27

So total number of girl vs boys matches =(18×25+27×24)=1098

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## Slot -1 – Quantitative Aptitude – Modern Maths – Permutation and Combination – How many numbers with two or more

How many numbers with two or more digits can be formed with the digits 1,2,3,4,5,6,7,8,9, so that in every such number, each digit is used at most once and the digits appear in the ascending order?

Solution :

As all given digits are different natural numbers and we have to use them at most once, ways of arrangements for each set of digits will be exactly once.
Thus Total such numbers = 9(baseC2) + 9(baseC3) +9(baseC4) +9(baseC5) +9(baseC6) + 9(baseC7) +9(baseC8) +9(baseC9) = 2^9 – 2 = 510

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## Slot -1 – Quantitative Aptitude – Modern Maths – Set Theory – Each of 74 students in a class studies

Each of 74 students in a class studies at least one of the three subjects H, E and P. Ten students study all three subjects, while twenty study H and E, but not P. Every student who studies P also studies H or E or both. If the number of students studying H equals that studying E, then the number of students studying H is?

Solution:

On the basis of given information , following venn-diagram can be made Now as the number of students studying H equals that studying E, so remaining (74 – 10 – 20 = 44) students must be equally distributed to E and H . So number students studying H = 10+20+ 44/2 = 52

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